2
$\begingroup$

In my assignment on quantum mechanics, I have encountered a sum $$\sum_{n\neq 1}\langle nlm|z|100\rangle$$ for hydrogen states $|nlm\rangle$. From the selection rule, we know that $l$ can only be $1$, $m$ can only be $0$. The problem will be reduced to $$\sum_{n\neq 1}\langle n10|z|100\rangle$$ but still, in the coordinate basis, the inner product integral is difficult to evaluate for large $n$ (not even talk about the sum). So, is there any good idea on how to evaluate this sum?

$\endgroup$
1
  • $\begingroup$ Are you sure that you need to evaluate the infinite sum? It seems like a Herculean task for an assignment. $\endgroup$ Commented Oct 24, 2016 at 2:20

1 Answer 1

1
$\begingroup$

Since $z = r\cos\theta$, the angular integral can be evaluated and the remaining difficulty is in the matrix elements $\langle nl|r|10\rangle$.

The paper (unfortunately paywalled)

presents various recurrence relations for matrix elements of the operator $r$. In particular equation (5.7) of the paper is

$$C^{-1}_{n,l-1} 2l \langle n' l |r|n,l-1 \rangle= C^{-1}_{n',l} (2l+1) \langle n', l+1|r|nl\rangle + C^{-1}_{nl} \langle n' l |r|n, l+1 \rangle \tag{5.7}$$ where $$C_{nl} = [1/(l+1)^2 - 1/n^2]^{-1/2} \tag{2.17}.$$

Even so, the infinite sum certainly seems intractable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.